Let be a compact interval of positive length (thus ). Recall that a function is said to be *differentiable* at a point if the limit

exists. In that case, we call the *strong derivative*, *classical derivative*, or just *derivative* for short, of at . We say that is *everywhere differentiable*, or differentiable for short, if it is differentiable at all points , and *differentiable almost everywhere* if it is differentiable at almost every point . If is differentiable everywhere and its derivative is continuous, then we say that is *continuously differentiable*.

Remark 1Much later in this sequence, when we cover the theory of distributions, we will see the notion of aweak derivativeordistributional derivative, which can be applied to a much rougher class of functions and is in many ways more suitable than the classical derivative for doing “Lebesgue” type analysis (i.e. analysis centred around the Lebesgue integral, and in particular allowing functions to be uncontrolled, infinite, or even undefined on sets of measure zero). However, for now we will stick with the classical approach to differentiation.

Exercise 2If is everywhere differentiable, show that is continuous and is measurable. If is almost everywhere differentiable, show that the (almost everywhere defined) function is measurable (i.e. it is equal to an everywhere defined measurable function on outside of a null set), but give an example to demonstrate that need not be continuous.

Exercise 3Give an example of a function which is everywhere differentiable, but not continuously differentiable. (Hint:choose an that vanishes quickly at some point, say at the origin , but which also oscillates rapidly near that point.)

In single-variable calculus, the operations of integration and differentiation are connected by a number of basic theorems, starting with Rolle’s theorem.

Theorem 4 (Rolle’s theorem)Let be a compact interval of positive length, and let be a differentiable function such that . Then there exists such that .

*Proof:* By subtracting a constant from (which does not affect differentiability or the derivative) we may assume that . If is identically zero then the claim is trivial, so assume that is non-zero somewhere. By replacing with if necessary, we may assume that is positive somewhere, thus . On the other hand, as is continuous and is compact, must attain its maximum somewhere, thus there exists such that for all . Then must be positive and so cannot equal either or , and thus must lie in the interior. From the right limit of (1) we see that , while from the left limit we have . Thus and the claim follows.

Remark 5Observe that the same proof also works if is only differentiable in the interior of the interval , so long as it is continuous all the way up to the boundary of .

Exercise 6Give an example to show that Rolle’s theorem can fail if is merely assumed to be almost everywhere differentiable, even if one adds the additional hypothesis that is continuous. This example illustrates that everywhere differentiability is a significantly stronger property than almost everywhere differentiability. We will see further evidence of this fact later in these notes; there are many theorems that assert in their conclusion that a function is almost everywhere differentiable, but few that manage to concludeeverywheredifferentiability.

Remark 7It is important to note that Rolle’s theorem only works in the real scalar case when is real-valued, as it relies heavily on the least upper bound property for the domain . If, for instance, we consider complex-valued scalar functions , then the theorem can fail; for instance, the function defined by vanishes at both endpoints and is differentiable, but its derivative is never zero. (Rolle’s theorem does imply that the real and imaginary parts of the derivative both vanish somewhere, but the problem is that they don’tsimultaneouslyvanish at the same point.) Similar remarks to functions taking values in a finite-dimensional vector space, such as .

One can easily amplify Rolle’s theorem to the mean value theorem:

Corollary 8 (Mean value theorem)Let be a compact interval of positive length, and let be a differentiable function. Then there exists such that .

*Proof:* Apply Rolle’s theorem to the function .

Remark 9As Rolle’s theorem is only applicable to real scalar-valued functions, the more general mean value theorem is also only applicable to such functions.

Exercise 10 (Uniqueness of antiderivatives up to constants)Let be a compact interval of positive length, and let and be differentiable functions. Show that for every if and only if for some constant and all .

We can use the mean value theorem to deduce one of the fundamental theorems of calculus:

Theorem 11 (Second fundamental theorem of calculus)Let be a differentiable function, such that is Riemann integrable. Then the Riemann integral of is equal to . In particular, we have whenever is continuously differentiable.

*Proof:* Let . By the definition of Riemann integrability, there exists a finite partition such that

for every choice of .

Fix this partition. From the mean value theorem, for each one can find such that

and thus by telescoping series

Since was arbitrary, the claim follows.

Remark 12Even though the mean value theorem only holds for real scalar functions, the fundamental theorem of calculus holds for complex or vector-valued functions, as one can simply apply that theorem to each component of that function separately.

Of course, we also have the other half of the fundamental theorem of calculus:

Theorem 13 (First fundamental theorem of calculus)Let be a compact interval of positive length. Let be a continuous function, and let be the indefinite integral . Then is differentiable on , with derivative for all . In particular, is continuously differentiable.

*Proof:* It suffices to show that

for all , and

for all . After a change of variables, we can write

for any and any sufficiently small , or any and any sufficiently small . As is continuous, the function converges uniformly to on as (keeping fixed). As the interval is bounded, thus converges to , and the claim follows.

Corollary 14 (Differentiation theorem for continuous functions)Let be a continuous function on a compact interval. Then we havefor all ,

for all , and thus

for all .

In these notes we explore the question of the extent to which these theorems continue to hold when the differentiability or integrability conditions on the various functions are relaxed. Among the results proven in these notes are

- The Lebesgue differentiation theorem, which roughly speaking asserts that Corollary 14 continues to hold for
*almost*every if is merely absolutely integrable, rather than continuous; - A number of
*differentiation theorems*, which assert for instance that monotone, Lipschitz, or bounded variation functions in one dimension are almost everywhere differentiable; and - The second fundamental theorem of calculus for absolutely continuous functions.

The material here is loosely based on Chapter 3 of Stein-Shakarchi.

** — 1. The Lebesgue differentiation theorem in one dimension — **

The main objective of this section is to show

Theorem 15 (Lebesgue differentiation theorem, one-dimensional case)Let be an absolutely integrable function, and let be the definite integral . Then is continuous and almost everywhere differentiable, and for almost every .

This can be viewed as a variant of Corollary 14; the hypotheses are weaker because is only assumed to be absolutely integrable, rather than continuous (and can live on the entire real line, and not just on a compact interval); but the conclusion is weaker too, because is only found to be almost everywhere differentiable, rather than everywhere differentiable. (But such a relaxation of the conclusion is necessary at this level of generality; consider for instance the example when .)

The continuity is an easy exercise:

Exercise 16Let be an absolutely integrable function, and let be the definite integral . Show that is continuous.

The main difficulty is to show that for almost every . This will follow from

Theorem 17 (Lebesgue differentiation theorem, second formulation)Let be an absolutely integrable function. Then

We will just prove the first fact (2); the second fact (3) is similar (or can be deduced from (2) by replacing with the reflected function .

We are taking to be complex valued, but it is clear from taking real and imaginary parts that it suffices to prove the claim when is real-valued, and we shall thus assume this for the rest of the argument.

The conclusion (2) we want to prove is a convergence theorem – an assertion that for all functions in a given class (in this case, the class of absolutely integrable functions ), a certain sequence of linear expressions (in this case, the right averages ) converge in some sense (in this case, pointwise almost everywhere) to a specified limit (in this case, ). There is a general and very useful argument to prove such convergence theorems, known as the *density argument*. This argument requires two ingredients, which we state informally as follows:

- A verification of the convergence result for some “dense subclass” of “nice” functions , such as continuous functions, smooth functions, simple functions, etc.. By “dense”, we mean that a general function in the original class can be approximated to arbitrary accuracy in a suitable sense by a function in the nice subclass.
- A quantitative estimate that upper bounds the maximal fluctuation of the linear expressions in terms of the “size” of the function (where the precise definition of “size” depends on the nature of the approximation in the first ingredient).

Once one has these two ingredients, it is usually not too hard to put them together to obtain the desired convergence theorem for general functions (not just those in the dense subclass). We illustrate this with a simple example:

Proposition 19 (Translation is continuous in )Let be an absolutely integrable function, and for each , let be the shifted functionThen converges in norm to as , thus

*Proof:* We first verify this claim for a dense subclass of , namely the functions which are continuous and compactly supported (i.e. they vanish outside of a compact set). Such functions are continuous, and thus converges uniformly to as . Furthermore, as is compactly supported, the support of stays uniformly bounded for in a bounded set. From this we see that also converges to in norm as required.

Next, we observe the quantitative estimate

for any . This follows easily from the triangle inequality

together with the translation invariance of the Lebesgue integral:

Now we put the two ingredients together. Let be absolutely integrable, and let be arbitrary. Applying Littlewood’s second principle (Theorem 15 from Notes 2) to the absolutely integrable function , we can find a continuous, compactly supported function such that

Applying (4), we conclude that

which we rearrange as

By the dense subclass result, we also know that

for all sufficiently close to zero. From the triangle inequality, we conclude that

for all sufficiently close to zero, and the claim follows.

Remark 20In the above application of the density argument, we proved the required quantitative estimate directly for all functions in the original class of functions. However, it is also possible to use the density argument a second time and initially verify the quantitative estimate just for functions in a nice subclass (e.g. continuous functions of compact support). In many cases, one can then extend that estimate to the general case by using tools such as Fatou’s lemma, which are particularly suited for showing that upper bound estimates are preserved with respect to limits.

Exercise 21Let , be Lebesgue measurable functions such that is absolutely integrable and is essentially bounded (i.e. bounded outside of a null set). Show that the convolution defined by the formulais well-defined (in the sense that the integrand on the right-hand side is absolutely integrable) and that is a bounded, continuous function.

The above exercise is illustrative of a more general intuition, which is that convolutions tend to be *smoothing* in nature; the convolution of two functions is usually at least as regular as, and often more regular than, either of the two factors .

This smoothing phenomenon gives rise to an important fact, namely the Steinhaus theorem:

Exercise 22 (Steinhaus theorem)Let be a Lebesgue measurable set of positive measure. Show that the set contains an open neighbourhood of the origin. (Hint:reduce to the case when is bounded, and then apply the previous exercise to the convolution , where .)

Exercise 23Ahomomorphismis a map with the property that for all .

- Show that all measurable homomorphisms are continuous. (
Hint:for any disk centered at the origin in the complex plane, show that has positive measure for at least one , and then use the Steinhaus theorem from the previous exercise.)- Show that is a measurable homomorphism if and only if it takes the form for all and some complex coefficients . (
Hint:first establish this for rational , and then use the previous part of this exercise.)- (For readers familiar with Zorn’s lemma) Show that there exist homomorphisms which are not of the form in the previous exercise. (Hint: view (or ) as a vector space over the rationals , and use the fact (from Zorn’s lemma) that every vector space – even an infinite-dimensional one – has at least one basis.) This gives an alternate construction of a non-measurable set to that given in previous notes.

Remark 24One drawback with the density argument is it gives convergence results which arequalitativerather than quantitative – there is no explicit bound on the rate of convergence. For instance, in Proposition 19, we know that for any , there exists such that whenever , but we do not know exactly how depends on and . Actually, the proof does eventually give such a bound, but it depends on “how measurable” the function is, or more precisely how “easy” it is to approximate by a “nice” function. To illustrate this issue, let’s work in one dimension and consider the function , where is a large integer. On the one hand, is bounded in the norm uniformly in : (indeed, the left-hand side is equal to ). On the other hand, it is not hard to see that for some absolute constant . Thus, if one force to drop below , one has to make at most from the origin. Making large, we thus see that the rate of convergence of to zero can be arbitrarily slow, even though is bounded in . The problem is that as gets large, it becomes increasingly difficult to approximate well by a “nice” function, by which we mean a uniformly continuous function with a reasonable modulus of continuity, due to the increasingly oscillatory nature of . See this blog post for some further discussion of this issue, and what quantitative substitutes are available for such qualitative results.

Now we return to the Lebesgue differentiation theorem, and apply the density argument. The dense subclass result is already contained in Corollary 14, which asserts that (2) holds for all continuous functions . The quantitative estimate we will need is the following special case of the Hardy-Littlewood maximal inequality:

Lemma 25 (One-sided Hardy-Littlewood maximal inequality)Let be an absolutely integrable function, and let . Then

We will prove this lemma shortly, but let us first see how this, combined with the dense subclass result, will give the Lebesgue differentiation theorem. Let be absolutely integrable, and let be arbitrary. Then by Littlewood’s second principle, we can find a function which is continuous and compactly supported, with

Applying the one-sided Hardy-Littlewood maximal inequality, we conclude that

In a similar spirit, from Markov’s inequality we have

By subadditivity, we conclude that for all outside of a set of measure at most , one has both

Now let . From the dense subclass result (Corollary 14) applied to the continuous function , we have

whenever is sufficiently close to . Combining this with (5), (6), and the triangle inequality, we conclude that

for all sufficiently close to zero. In particular we have

for all outside of a set of measure . Keeping fixed and sending to zero, we conclude that

for almost every . If we then let go to zero along a countable sequence (e.g. for ), we conclude that

for almost every , and the claim follows.

The only remaining task is to establish the one-sided Hardy-Littlewood maximal inequality. We will do so by using the rising sun lemma:

Lemma 26 (Rising sun lemma)Let be a compact interval, and let be a continuous function. Then one can find an at most countable family of disjoint non-empty open intervals in with the following properties:

- For each , either , or else and .
- If does not lie in any of the intervals , then one must have for all .

Remark 27To explain the name “rising sun lemma”, imagine the graph of as depicting a hilly landscape, with the sun shining horizontally from the rightward infinity (or rising from the east, if you will). Those for which are the locations on the landscape which are illuminated by the sun. The intervals then represent the portions of the landscape that are in shadow.

This lemma is proven using the following basic fact:

Exercise 28Show that any open subset of can be written as the union of at most countably many disjoint non-empty open intervals, whose endpoints lie outside of . (Hint:first show that every in is contained in a maximal open subinterval of , and that these maximal open subintervals are disjoint, with each such interval containing at least one rational number.)

*Proof:* (Proof of rising sun lemma) Let be the set of all such that for at least one . As is continuous, is open, and so is the union of at most countably many disjoint non-empty open intervals , with the endpoints lying outside of .

The second conclusion of the rising sun lemma is clear from construction, so it suffices to establish the first. Suppose first that is such that . As the endpoint does not lie in , we must have for all ; similarly we have for all . In particular we have . By the continuity of , it will then suffice to show that for all .

Suppose for contradiction that there was with . Let , then is a closed set that contains but is disjoint from , since , and for all , we see that cannot exceed , and thus lies in , but this contradicts the fact that is the supremum of .

The case when is similar and is left to the reader; the only difference is that we can no longer assert that for all , and so do not have the upper bound .

Now we can prove the one-sided Hardy-Littlewood maximal inequality. By upwards monotonicity, it will suffice to show that

for any compact interval . By modifying by an epsilon, we may replace the non-strict inequality here with strict inequality:

Fix . We apply the rising sun lemma to the function defined as

By Lemma 16, is continuous, and so we can find an at most countable sequence of intervals with the properties given by the rising sun lemma. From the second property of that lemma, we observe that

since the property can be rearranged as . By countable additivity, we may thus upper bound the left-hand side of (7) by . On the other hand, since , we have

and thus

As the are disjoint intervals in , we may apply monotone convergence and monotonicity to conclude that

and the claim follows.

Exercise 29 (Two-sided Hardy-Littlewood maximal inequality)Let be an absolutely integrable function, and let . Show thatwhere the supremum ranges over all intervals of positive length that contain .

Exercise 30 (Rising sun inequality)Let be an absolutely integrable function, and let be the one-sided signed Hardy-Littlewood maximal functionEstablish the

rising sun inequalityfor all real (note here that we permit to be zero or negative), and show that this inequality implies Lemma 36. (

Hint:First do the case, by invoking the rising sun lemma.) See these lecture notes for some further discussion of inequalities of this type, and applications to ergodic theory (and in particular the maximal ergodic theorem).

Exercise 31Show that the left and right-hand sides in Exercise 30 are in fact equal when . (Hint:one may first wish to try this in the case when has compact support, in which case one can apply the rising sun lemma to a sufficiently large interval containing the support of .)

** — 2. The Lebesgue differentiation theorem in higher dimensions — **

Now we extend the Lebesgue differentiation theorem to higher dimensions. Theorem 15 does not have an obvious high-dimensional analogue, but Theorem 17 does:

Theorem 32 (Lebesgue differentiation theorem in high dimensions)Let be an absolutely integrable function. Then for almost every , one haswhere is the open ball of radius centred at .

From the triangle inequality we see that

so we see that the first conclusion of Theorem 32 implies the second. A point for which (8) holds is called a *Lebesgue point* of ; thus, for an absolutely integrable function , almost every point in will be a Lebesgue point for .

Exercise 33Call a function locally integrable if, for every , there exists an open neighbourhood of on which is absolutely integrable.

- Show that is locally integrable if and only if .
- Show that Theorem 32 implies a generalisation of itself in which the condition of absolute integrability of is weakened to local integrability.

Exercise 34For each , let be a subset of with the property that for some independent of . Show that if is locally integrable, and is a Lebesgue point of , then

To prove Theorem 32, we use the density argument. The dense subclass case is easy:

Exercise 35Show that Theorem 32 holds whenever is continuous.

The quantitative estimate needed is the following:

Theorem 36 (Hardy-Littlewood maximal inequality)Let be an absolutely integrable function, and let . Thenfor some constant depending only on .

Remark 37The expression is known as the Hardy-Littlewood maximal function of , and is often denoted . It is an important function in the field of (real-variable) harmonic analysis.

Exercise 38Use the density argument to show that Theorem 36 implies Theorem 32.

In the one-dimensional case, this estimate was established via the rising sun lemma. Unfortunately, that lemma relied heavily on the ordered nature of , and does not have an obvious analogue in higher dimensions. Instead, we will use the following covering lemma. Given an open ball in and a real number , we write for the ball with the same centre as , but times the radius. (Note that this is slightly different from the set – why?) Note that for any open ball and any .

Lemma 39 (Vitali-type covering lemma)Let be a finite collection of open balls in (not necessarily disjoint). Then there exists a subcollection ofdisjointballs in this collection, such that

*Proof:* We use a greedy algorithm argument, selecting the balls to be as large as possible while remaining disjoint. More precisely, we run the following algorithm:

- Step 0. Initialise (so that, initially, there are no balls in the desired collection).
- Step 1. Look at all the balls that do not already intersect one of the (which, initially, will be all the balls ). If there are no such balls, STOP. Otherwise, go on to Step 2.
- Step 2. Locate the largest ball that does not already intersect one of the . (If there are multiple largest balls with exactly the same radius, break the tie arbitrarily.) Add this ball to the collection by setting and then incrementing to . Then return to Step 1.

Note that at each iteration of this algorithm, the number of available balls amongst the drops by at least one (since each ball selected certainly intersects itself and so cannot be selected again). So this algorithm terminates in finite time. It is also clear from construction that the are a subcollection of the consisting of disjoint balls. So the only task remaining is to verify that (9) holds at the completion of the algorithm, i.e. to show that each ball in the original collection is covered by the triples of the subcollection.

For this, we argue as follows. Take any ball in the original collection. Because the algorithm only halts when there are no more balls that are disjoint from the , the ball must intersect at least one of the balls in the subcollection. Let be the first ball with this property, thus is disjoint from , but intersects . Because was chosen to be largest amongst all balls that did not intersect , we conclude that the radius of cannot exceed that of . From the triangle inequality, this implies that , and the claim follows.

Exercise 40Technically speaking, the above algorithmic argument was not phrased in the standard language of formal mathematical deduction, because in that language, any mathematical object (such as the natural number ) can only be defined once, and not redefined multiple times as is done in most algorithms. Rewrite the above argument in a way that avoids redefining any variable. (Hint:introduce a “time” variable , and recursively construct families of balls that represent the outcome of the above algorithm after iterations (or iterations, if the algorithm halted at some previous time ). For this particular algorithm, there are also moread hocapproaches that exploit the relatively simple nature of the algorithm to allow for a less notationally complicated construction.) More generally, it is possible to use this time parameter trick to convert any construction involving a provably terminating algorithm into a construction that does not redefine any variable. (It is however dangerous to work with any algorithm that has an infinite run time, unless one has a suitably strong convergence result for the algorithm that allows one to take limits, either in the classical sense or in the more general sense of jumping to limit ordinals; in the latter case, one needs to use transfinite induction in order to ensure that the use of such algorithms is rigorous.)

Remark 41The actual Vitali covering lemma is slightly different to this one, as the linked Wikipedia page shows. Actually there is a family of related covering lemmas which are useful for a variety of tasks in harmonic analysis, see for instance this book by de Guzmán for further discussion.

Now we can prove the Hardy-Littlewood inequality, which we will do with the constant . It suffices to verify the claim with strict inequality,

as the non-strict case then follows by perturbing slightly and then taking limits.

Fix and . By inner regularity, it suffices to show that

whenever is a compact set that is contained in .

By construction, for every , there exists an open ball such that

By compactness of , we can cover by a finite number of such balls. Applying the Vitali-type covering lemma, we can find a subcollection of disjoint balls such that

By (10), on each ball we have

summing in and using the disjointness of the we conclude that

Since the cover , we obtain Theorem 36 as desired.

Exercise 42Improve the constant in the Hardy-Littlewood maximal inequality to . (Hint:observe that with the construction used to prove the Vitali covering lemma, thecentresof the balls are contained in and not just in . To exploit this observation one may need to first create an epsilon of room, as the centers are not by themselves sufficient to cover the required set.)

Remark 43The optimal value of is not known in general, although a fairly recent result of Melas gives the surprising conclusion that the optimal value of is . It is known that grows at most linearly in , thanks to a result of Stein and Strömberg, but it is not known if is bounded in or grows as . See this blog post for some further discussion.

Exercise 44 (Dyadic maximal inequality)If is an absolutely integrable function, establish the dyadic Hardy-Littlewood maximal inequalitywhere the supremum ranges over all dyadic cubes that contain . (

Hint:the nesting property of dyadic cubes will be useful when it comes to the covering lemma stage of the argument, much as it was in Exercise 8 of Notes 1.)

Exercise 45 (Besicovich covering lemma in one dimension)Let be a finite family of open intervals in (not necessarily disjoint). Show that there exist a subfamily of intervals such that

- ; and
- Each point is contained in at most two of the .
(

Hint:First refine the family of intervals so that no interval is contained in the union of the the other intervals. At that point, show that it is no longer possible for a point to be contained in three of the intervals.) There is a variant of this lemma that holds in higher dimensions, known as the Besicovitch covering lemma.

Exercise 46Let be a Borel measure (i.e. a countably additive measure on the Borel -algebra) on , such that for every interval of positive length. Assume that isinner regular, in the sense that for every Borel measurable set . (As it turns out, from the theory of Radon measures, all locally finite Borel measures have this property, but we will not prove this here; see Exercise 12 of these notes.) Establish the Hardy-Littlewood maximal inequalityfor any absolutely integrable function , where the supremum ranges over all open intervals that contain . Note that this essentially generalises Exercise 29, in which is replaced by Lebesgue measure. (

Hint:Repeat the proof of the usual Hardy-Littlewood maximal inequality, but use the Besicovich covering lemma in place of the Vitaly-type covering lemma. Why do we need the former lemma here instead of the latter?)

Exercise 47 (Cousin’s theorem)Prove Cousin’s theorem: given any function on a compact interval of positive length, there exists a partition with , together with real numbers for each and . (Hint:use the Heine-Borel theorem, which asserts that any open cover of has a finite subcover, followed by the Besicovitch covering lemma.) This theorem is useful in a variety of applications related to the second fundamental theorem of calculus, as we shall see below. The positive function is known as agauge function.

Now we turn to consequences of the Lebesgue differentiation theorem. Given a Lebesgue measurable set , call a point a *point of density* for if as . Thus, for instance, if , then every point in (including the boundary point ) is a point of density for , but the endpoints (as well as the exterior of ) are not points of density. One can think of a point of density as being an “almost interior” point of ; it is not necessarily the case that one can fit an small ball centred at inside of , but one can fit *most* of that small ball inside .

Exercise 48If is Lebesgue measurable, show that almost every point in is a point of density for , and almost every point in the complement of is not a point of density for .

Exercise 49Let be a measurable set of positive measure, and let .

- Using Exercise 34 and Exercise 48, show that there exists a cube of positive sidelength such that .
- Give an alternate proof of the above claim that avoids the Lebesgue differentiation theorem. (
Hint:reduce to the case when is bounded, then approximate by an almost disjoint union of cubes.)- Use the above result to give an alternate proof of the Steinhaus theorem (Exercise 22).

Of course, one can replace cubes here by other comparable shapes, such as balls. (Indeed, a good principle to adopt in analysis is that cubes and balls are “equivalent up to constants”, in that a cube of some sidelength can be contained in a ball of comparable radius, and vice versa. This type of mental equivalence is analogous to, though not identical with, the famous dictum that a topologist cannot distinguish a doughnut from a coffee cup.)

Exercise 50

- Give an example of a compact set of positive measure such that for every interval of positive length. (
Hint:first construct an open dense subset of of measure strictly less than .)- Give an example of a measurable set such that for every interval of positive length. (
Hint:first work in a bounded interval, such as . The complement of the set in the first example is the union of at most countably many open intervals, thanks to Exercise 28. Now fill in these open intervals and iterate.)

Exercise 51 (Approximations to the identity)Define agood kernelto be a measurable function which is non-negative, radial (which means that there is a function such that ), radially non-increasing (so that is a non-increasing function), and has total mass equal to . The functions for are then said to be agood family of approximations to the identity.

- Show that the heat kernels and Poisson kernels are good families of approximations to the identity, if the constant is chosen correctly (in fact one has , but you are not required to establish this). (Note that we have modified the usual formulation of the heat kernel by replacing with in order to make it conform to the notational conventions used in this exercise.)
- Show that if is a good kernel, then
for some constants depending only on . (

Hint:compare with such “horizontal wedding cake” functions as .)- Establish the quantitative upper bound
for any absolutely integrable function and some constant depending only on .

- Show that if is absolutely integrable and is a Lebesgue point of , then the convolution
converges to as . (

Hint:split as the sum of and .) In particular, converges pointwise almost everywhere to .

** — 3. Almost everywhere differentiability — **

As we see in undergraduate real analysis, not every continuous function is differentiable, with the standard example being the absolute value function , which is continuous not differentiable at the origin . Of course, this function is still almost everywhere differentiable. With a bit more effort, one can construct continuous functions that are in fact nowhere differentiable:

Exercise 52 (Weierstrass function)Let be the function

- Show that is well-defined (in the sense that the series is absolutely convergent) and that is a bounded continuous function.
- Show that for every interval with and integer, one has for some absolute constant .
- Show that is not differentiable at any point . (
Hint:argue by contradiction and use the previous part of this exercise.) Note that it isnotenough to formally differentiate the series term by term and observe that the resulting series is divergent – why not?

The difficulty here is that a continuous function can still contain a large amount of *oscillation*, which can lead to breakdown of differentiability. However, if one can somehow limit the amount of oscillation present, then one can often recover a fair bit of differentiability. For instance, we have

Theorem 53 (Monotone differentiation theorem)Any function which is monotone (either monotone non-decreasing or monotone non-increasing) is differentiable almost everywhere.

Exercise 54Show that every monotone function is measurable.

To prove this theorem, we just treat the case when is monotone non-decreasing, as the non-increasing case is similar (and can be deduced from the non-decreasing case by replacing with ).

We also first focus on the case when is continuous, as this allows us to use the rising sun lemma. To understand the differentiability of , we introduce the four Dini derivatives of at :

- The upper right derivative ;
- The lower right derivative ;
- The upper left derivative ;
- The lower right derivative .

Regardless of whether is differentiable or not (or even whether is continuous or not), the four Dini derivatives always exist and take values in the extended real line . (If is only defined on an interval , rather than on the endpoints, then some of the Dini derivatives may not exist at the endpoints, but this is a measure zero set and will not impact our analysis.)

Exercise 55If is monotone, show that the four Dini derivatives of are measurable. (Hint:the main difficulty is to reformulate the derivatives so that ranges over a countable set rather than an uncountable one.)

A function is differentiable at precisely when the four derivatives are equal and finite:

We also have the trivial inequalities

If is non-decreasing, all these quantities are non-negative, thus

The one-sided Hardy-Littlewood maximal inequality has an analogue in this setting:

Lemma 56 (One-sided Hardy-Littlewood inequality)Let be a continuous monotone non-decreasing function, and let . Then we haveSimilarly for the other three Dini derivatives of .

If is not assumed to be continuous, then we have the weaker inequality

for some absolute constant .

Remark 57Note that if one naively applies the fundamental theorems of calculus, one canformallysee that the first part of Lemma 56 is equivalent to Lemma 36. We cannot however use this argument rigorously because we have not established the necessary fundamental theorems of calculus to do this. Nevertheless, we can borrow theproofof Lemma 36 without difficulty to use here, and this is exactly what we will do.

*Proof:* We just prove the continuous case and leave the discontinuous case as an exercise.

It suffices to prove the claim for ; by reflection (replacing with , and with ), the same argument works for , and then this trivially implies the same inequalities for and . By modifying by an epsilon, and dropping the endpoints from as they have measure zero, it suffices to show that

We may apply the rising sun lemma (Lemma 26) to the continuous function . This gives us an at most countable family of intervals in , such that for each , and such that whenever and lies outside of all of the .

Observe that if , and for all , then . Thus we see that the set is contained in the union of the , and so by countable additivity

But we can rearrange the inequality as . From telescoping series and the monotone nature of we have (this is easiest to prove by first working with a finite subcollection of the intervals , and then taking suprema), and the claim follows.

The discontinuous case is left as an exercise.

Exercise 58Prove Lemma 56 in the discontinuous case. (Hint:the rising sun lemma is no longer available, but one can use either the Vitali-type covering lemma (which will give ) or the Besicovitch lemma (which will give ), by modifying the proof of Theorem 36.

Sending in the above lemma (cf. Exercise 18 from Notes 2), and then sending to , we conclude as a corollary that all the four Dini derivatives of a continuous monotone non-decreasing function are finite almost everywhere. So to prove Theorem 53 for continuous monotone non-decreasing functions, it suffices to show that (11) holds for almost every . In view of the trivial inequalities, it suffices to show that and for almost every . We will just show the first inequality, as the second follows by replacing with its reflection . It will suffice to show that for every pair of real numbers, the set

is a null set, since by letting range over rationals with and taking countable unions, we would conclude that the set is a null set (recall that the Dini derivatives are all non-negative when is non-decreasing), and the claim follows.

Clearly is a measurable set. To prove that it is null, we will establish the following estimate:

Lemma 59 ( has density less than one)For any interval and any , one has .

Indeed, this lemma implies that has no points of density, which by Exercise 48 forces to be a null set.

*Proof:* We begin by applying the rising sun lemma to the function on ; the large number of negative signs present here is needed in order to properly deal with the lower left Dini derivative . This gives an at most countable family of disjoint intervals in , such that for all , and such that whenever and lies outside of all of the . Observe that if , and for all , then . Thus we see that is contained inside the union of the intervals . On the other hand, from the first part of Lemma 56 we have

But we can rearrange the inequality as . From countable additivity, one thus has

But the are disjoint inside , so from countable additivity again, we have , and the claim follows.

Remark 60Note if was not assumed to be continuous, then one would lose a factor of here from the second part of Lemma 56, and one would then be unable to prevent from being up to times as large as . So sometimes, even when all one is seeking is a qualitative result such as differentiability, it is still important to keep track of constants. (But this is the exception rather than the rule: for a large portion of arguments in analysis, the constants are not terribly important.)

This concludes the proof of Theorem 53 in the continuous monotone non-decreasing case. Now we work on removing the continuity hypothesis (which was needed in order to make the rising sun lemma work properly). If we naively try to run the density argument as we did in previous sections, then (for once) the argument does *not* work very well, as the space of continuous monotone functions are not sufficiently dense in the space of all monotone functions in the relevant sense (which, in this case, is in the total variation sense, which is what is needed to invoke such tools as Lemma 56.). To bridge this gap, we have to supplement the continuous monotone functions with another class of monotone functions, known as the *jump functions*.

Definition 61 (Jump function)Abasic jump functionis a function of the formfor some real numbers and ; we call the

point of discontinuityfor and thefraction. Observe that such functions are monotone non-decreasing, but have a discontinuity at one point. Ajump functionis any absolutely convergent combination of basic jump functions, i.e. a function of the form , where ranges over an at most countable set, each is a basic jump function, and the are positivereals with . If there are only finitely many involved, we say that is apiecewise constant jump function.

Thus, for instance, if is any enumeration of the rationals, then is a jump function.

Clearly, all jump functions are monotone non-decreasing. From the absolute convergence of the we see that every jump function is the uniform limit of piecewise constant jump functions, for instance is the uniform limit of . One consequence of this is that the points of discontinuity of a jump function are precisely those of the individual summands , i.e. of the points where each jumps.

The key fact is that these functions, together with the continuous monotone functions, essentially generate all monotone functions, at least in the bounded case:

Lemma 62 (Continuous-singular decomposition for monotone functions)Let be a monotone non-decreasing function.

- The only discontinuities of are jump discontinuities. More precisely, if is a point where is discontinuous, then the limits and both exist, but are unequal, with .
- There are at most countably many discontinuities of .
- If is bounded, then can be expressed as the sum of a continuous monotone non-decreasing function and a jump function .

Remark 63This decomposition is part of the more general Lebesgue decomposition, which we will discuss later in this course.

*Proof:* By monotonicity, the limits and always exist, with for all . This gives 1.

By 1., whenever there is a discontinuity of , there is at least one rational number strictly between and , and from monotonicity, each rational number can be assigned to at most one discontinuity. This gives 2.

Now we prove 3. Let be the set of discontinuities of , thus is at most countable. For each , we define the jump , and the fraction . Thus

Note that is the measure of the interval . By monotonicity, these intervals are disjoint; by the boundedness of , their union is bounded. By countable additivity, we thus have , and so if we let be the basic jump function with point of discontinuity and fraction , then the function

is a jump function.

As discussed previously, is discontinuous only at , and for each one easily checks that

where , and . We thus see that the difference is continuous. The only remaining task is to verify that is monotone non-decreasing, thus we need

for all . But the left-hand side can be rewritten as . As each is the measure of the interval , and these intervals for are disjoint and lie in , the claim follows from countable additivity.

Exercise 64Show that the decomposition of a bounded monotone non-decreasing function into continuous and jump components given by the above lemma is unique.

Exercise 65Find a suitable generalisation of the notion of a jump function that allows one to extend the above decomposition to unbounded monotone functions, and then prove this extension. (Hint:the notion to shoot for here is that of a “locally jump function”.)

Now we can finish the proof of Theorem 53. As noted previously, it suffices to prove the claim for monotone non-decreasing functions. As differentiability is a local condition, we can easily reduce to the case of *bounded* monotone non-decreasing functions, since to test differentiability of a monotone non-decreasing function in any compact interval we may replace by the bounded monotone non-decreasing function with no change in the differentiability in (except perhaps at the endpoints , but these form a set of measure zero). As we have already proven the claim for continuous functions, it suffices by Lemma 62 (and linearity of the derivative) to verify the claim for jump functions.

Now, finally, we are able to use the density argument, using the piecewise constant jump functions as the dense subclass, and using the second part of Lemma 56 for the quantitative estimate; fortunately for us, the density argument does not particularly care that there is a loss of a constant factor in this estimate.

For piecewise constant jump functions, the claim is clear (indeed, the derivative exists and is zero outside of finitely many discontinuities). Now we run the density argument. Let be a bounded jump function, and let and be arbitrary. As every jump function is the uniform limit of piecewise constant jump functions, we can find a piecewise constant jump function such that for all . Indeed, by taking to be a partial sum of the basic jump functions that make up , we can ensure that is also a monotone non-decreasing function. Applying the second part of Lemma 56, we have

for some absolute constant , and similarly for the other four Dini derivatives. Thus, outside of a set of measure at most , all of the Dini derivatives of are less than . Since is almost everywhere differentiable, we conclude that outside of a set of measure at most , all the Dini derivatives of lie within of , and in particular are finite and lie within of each other. Sending to zero (holding fixed), we conclude that for almost every , the Dini derivatives of are finite and lie within of each other. If we then send to zero, we see that for almost every , the Dini derivatives of agree with each other and are finite, and the claim follows. This concludes the proof of Theorem 53.

Just as the integration theory of unsigned functions can be used to develop the integration theory of the absolutely convergent functions (see Notes 2), the differentiation theory of monotone functions can be used to develop a parallel differentiation theory for the class of functions of bounded variation:

Definition 66 (Bounded variation)Let be a function. Thetotal variation(or for short) of is defined to be the supremumwhere the supremum ranges over all finite increasing sequences of real numbers with ; this is a quantity in . We say that has

bounded variation(on ) if is finite. (In this case, is often written as or just .)Given any interval , we define the total variation of on as

thus the definition is the same, but the points are restricted to lie in . Thus for instance . We say that a function has

bounded variationon if is finite.

Exercise 67If is a monotone function, show that for any interval , and that has bounded variation on if and only if it is bounded.

Exercise 68For any functions , establish the triangle property and the homogeneity property for any . Also show that if and only if is constant.

Exercise 69If is a function, show that whenever .

Exercise 70

- Show that every function of bounded variation is bounded, and that the limits and , are well-defined.
- Give a counterexample of a bounded, continuous, compactly supported function that is not of bounded variation.

Exercise 71Let be an absolutely integrable function, and let be the indefinite integral . Show that is of bounded variation, and that . (Hint:the upper bound is relatively easy to establish. To obtain the lower bound, use the density argument.)

Much as an absolutely integrable function can be expressed as the difference of its positive and negative parts, a bounded variation function can be expressed as the difference of two bounded monotone functions:

Proposition 72A function is of bounded variation if and only if it is the difference of two bounded monotone functions.

*Proof:* It is clear from Exercises 67, 68 that the difference of two bounded monotone functions is bounded. Now define the *positive variation* of by the formula

It is clear from construction that this is a monotone increasing function, taking values between and , and is thus bounded. To conclude the proposition, it suffices to (by writing to show that is non-decreasing, or in other words to show that

If is negative then this is clear from the monotone non-decreasing nature of , so assume that . But then the claim follows because any sequence of real numbers can be extended by one or two elements by adding and , thus increasing the sum by at least .

Exercise 73Let be of bounded variation. Define the positive variation by (12), and the negative variation byEstablish the identities

and

for every interval , where , , and . (

Hint:The main difficulty comes from the fact that a partition that is good for need not be good for , and vice versa. However, this can be fixed by taking a good partition for and a good partition for and combining them together into a common refinement.)

From Proposition 72 and Theorem 53 we immediately obtain

Corollary 74 (BV differentiation theorem)Every bounded variation function is differentiable almost everywhere.

Exercise 75Call a functionlocally of bounded variationif it is of bounded variation on every compact interval . Show that every function that is locally of bounded variation is differentiable almost everywhere.

Exercise 76 (Lipschitz differentiation theorem, one-dimensional case)A function is said to beLipschitz continuousif there exists a constant such that for all ; the smallest with this property is known as theLipschitz constantof . Show that every Lipschitz continuous function is locally of bounded variation, and hence differentiable almost everywhere. Furthermore, show that the derivative , when it exists, is bounded in magnitude by the Lipschitz constant of .

Remark 77The same result is true in higher dimensions, and is known as the Radamacher differentiation theorem, but we will defer the proof of this theorem to subsequent notes, when we have the powerful tool of the Fubini-Tonelli theorem available, that is particularly useful for deducing higher-dimensional results in analysis from lower-dimensional ones.

Exercise 78A function is said to be convex if one has for all and . Show that if is convex, then it is continuous and almost everywhere differentiable, and its derivative is equal almost everywhere to a monotone non-decreasing function, and so is itself almost everywhere differentiable. (Hint:Drawing the graph of , together with a number of chords and tangent lines, is likely to be very helpful in providing visual intuition.) Thus we see that in some sense, convex functions are “almost everywhere twice differentiable”. Similar claims also hold for concave functions, of course.

** — 4. The second fundamental theorem of calculus — **

We are now finally ready to attack the second fundamental theorem of calculus in the cases where is not assumed to be continuously differentiable. We begin with the case when is monotone non-decreasing. From Theorem 53 (extending to the rest of the real line if needed), this implies that is differentiable almost everywhere in , so is defined a.e.; from monotonicity we see that is non-negative whenever it is defined. Also, an easy modification of Exercise 2 shows that is measurable.

One half of the second fundamental theorem is easy:

Proposition 79 (Upper bound for second fundamental theorem)Let be monotone non-decreasing (so that, as discussed above, is defined almost everywhere, is unsigned, and is measurable). ThenIn particular, is absolutely integrable.

*Proof:* It is convenient to extend to all of by declaring for and for , then is now a bounded monotone function on , and vanishes outside of . As is almost everywhere differentiable, the Newton quotients

converge pointwise almost everywhere to . Applying Fatou’s lemma (Corollary 16 of Notes 3), we conclude that

The right-hand side can be rearranged as

which can be rearranged further as

Since is equal to for the first integral and is at least for the second integral, this expression is at most

and the claim follows.

Exercise 80Show that any function of bounded variation has an (almost everywhere defined) derivative that is absolutely integrable.

In the Lipschitz case, one can do better:

Exercise 81 (Second fundamental theorem for Lipschitz functions)Let be Lipschitz continuous. Show that . (Hint:Argue as in the proof of Proposition 79, but use the dominated convergence theorem in place of Fatou’s lemma.)

Exercise 82 (Integration by parts formula)Let be Lipschitz continuous functions. Show that(

Hint:first show that the product of two Lipschitz continuous functions on is again Lipschitz continuous.)

Now we return to the monotone case. Inspired by the Lipschitz case, one may hope to recover equality in Proposition 79 for such functions . However, there is an important obstruction to this, which is that all the variation of may be concentrated in a set of measure zero, and thus undetectable by the Lebesgue integral of . This is most obvious in the case of a discontinuous monotone function, such as the (appropriately named) Heaviside function ; it is clear that vanishes almost everywhere, but is not equal to if and lie on opposite sides of the discontinuity at . In fact, the same problem arises for all jump functions:

Exercise 83Show that if is a jump function, then vanishes almost everywhere. (Hint:use the density argument, starting from piecewise constant jump functions and using Proposition 79 as the quantitative estimate.)

One may hope that jump functions – in which all the fluctuation is concentrated in a countable set – are the only obstruction to the second fundamental theorem of calculus holding for monotone functions, and that as long as one restricts attention to *continuous* monotone functions, that one can recover the second fundamental theorem. However, this is still not true, because it is possible for all the fluctuation to now be concentrated, not in a countable collection of jump discontinuities, but instead in an uncountable set of zero measure, such as the middle thirds Cantor set (Exercise 10 from Notes 1). This can be illustrated by the key counterexample of the Cantor function, also known as the *Devil’s staircase function*. The construction of this function is detailed in the exercise below.

Exercise 84 (Cantor function)Define the functions recursively as follows:

- Set for all .
- For each in turn, define

- Graph , , , and (preferably on a single graph).
- Show that for each , is a continuous monotone non-decreasing function with and . (
Hint:induct on .)- Show that for each , one has for each . Conclude that the converge uniformly to a limit . This limit is known as the
Cantor function.- Show that the Cantor function is continuous and monotone non-decreasing, with and .
- Show that if lies outside the middle thirds Cantor set (Exercise 10 from Notes 1), then is constant in a neighbourhood of , and in particular . Conclude that , so that the second fundamental theorem of calculus fails for this function.
- Show that for any digits . Thus the Cantor function, in some sense, converts base three expansions to base two expansions.
- Let be one of the intervals used in the cover of (see Exercise 10 from Notes 1), thus and . Show that is an interval of length , but is an interval of length .
- Show that is not differentiable at any element of the Cantor set .

Remark 85This example shows that the classical derivative of a function has some defects; it cannot “see” some of the variation of a continuous monotone function such as the Cantor function. Much later in this series, we will rectify this by introducing the concept of the weak derivative of a function, which despite the name, is more able than the strong derivative to detect this type of singular variation behaviour. (We will also encounter the Riemann-Stieltjes integral in later notes, which is another (closely related) way to capture all of the variation of a monotone function, and which is related to the classical derivative via the Lebesgue-Radon-Nikodym theorem.)

In view of this counterexample, we see that we need to add an additional hypothesis to the continuous monotone non-increasing function before we can recover the second fundamental theorem. One such hypothesis is absolute continuity. To motivate this definition, let us recall two existing definitions:

- A function is
*continuous*if, for every and , there exists a such that whenever is an interval of length at most that contains . - A function is
*uniformly continuous*if, for every , there exists a such that whenever is an interval of length at most .

Definition 86A function is said to beabsolutely continuousif, for every , there exists a such that whenever is a finite collection of disjoint intervals of total length at most .We define absolute continuity for a function defined on an interval similarly, with the only difference being that the intervals are of course now required to lie in the domain of .

The following exercise places absolute continuity in relation to other regularity properties:

- Show that every absolutely continuous function is uniformly continuous and therefore continuous.
- Show that every absolutely continuous function is of bounded variation on every compact interval . (
Hint:first show this is true for any sufficiently small interval.) In particular (by Exercise 75), absolutely continuous functions are differentiable almost everywhere.- Show that every Lipschitz continuous function is absolutely continuous.
- Show that the function is absolutely continuous, but not Lipschitz continuous, on the interval .
- Show that the Cantor function from Exercise 84 is continuous, monotone, and uniformly continuous, but
notabsolutely continuous, on .- If is absolutely integrable, show that the indefinite integral is absolutely continuous, and that is differentiable almost everywhere with for almost every .
- Show that the sum or product of two absolutely continuous functions on an interval remains absolutely continuous. What happens if we work on instead of on ?

Exercise 88

- Show that absolutely continuous functions map null sets to null sets, i.e. if is absolutely continuous and is a null set then is also a null set.
- Show that the Cantor function does
nothave this property.

For absolutely continuous functions, we can recover the second fundamental theorem of calculus:

Theorem 89 (Second fundamental theorem for absolutely continuous functions)Let be absolutely continuous. Then .

*Proof:* Our main tool here will be Cousin’s theorem (Exercise 47).

By Exercise 80, is absolutely integrable. By Exercise 8 of Notes 4, is thus uniformly integrable. Now let . By Exercise 11 of Notes 4, we can find such that whenever is a measurable set of measure at most . (Here we adopt the convention that vanishes outside of .) By making small enough, we may also assume from absolute continuity that whenever is a finite collection of disjoint intervals of total length at most .

Let be the set of points where is not differentiable, together with the endpoints , as well as the points where is not a Lebesgue point of . thus is a null set. By outer regularity (or the definition of outer measure) we can find an open set containing of measure . In particular, .

Now define a gauge function as follows.

- If , we define to be small enough that the open interval lies in .
- If , then is differentiable at and is a Lebesgue point of . We let be small enough that holds whenever , and such that whenever is an interval containing of length at most ; such a exists by the definition of differentiability, and of Lebesgue point. We rewrite these properties using big-O notation as and .

Applying Cousin’s theorem, we can find a partition with , together with real numbers for each and .

We can express as a telescoping series

To estimate the size of this sum, let us first consider those for which . Then, by construction, the intervals are disjoint in . By construction of , we thus have

and thus

Next, we consider those for which . By construction, for those we have

and

and thus

On the other hand, from construction again we have

and thus

Summing in , we conclude that

where is the union of all the with . By construction, this set is contained in and contains . Since , we conclude that

Putting everything together, we conclude that

Since was arbitrary, the claim follows.

Combining this result with Exercise 87, we obtain a satisfactory classification of the absolutely continuous functions:

Exercise 90Show that a function is absolutely continuous if and only if it takes the form for some absolutely integrable and a constant .

Exercise 91 (Compatibility of the strong and weak derivatives in the absolutely continuous case)Let be an absolutely continuous function, and let be a continuously differentiable function supported in a compact subset of . Show that .

Inspecting the proof of Theorem 89, we see that the absolute continuity was used primarily in two ways: firstly, to ensure the almost everywhere existence, and to control an exceptional null set . It turns out that one can achieve the latter control by making a different hypothesis, namely that the function is *everywhere* differentiable rather than merely almost everywhere differentiable. More precisely, we have

Proposition 92 (Second fundamental theorem of calculus, again)Let be a compact interval of positive length, let be a differentiable function, such that is absolutely integrable. Then the Lebesgue integral of is equal to .

*Proof:* This will be similar to the proof of Theorem 89, the one main new twist being that we need several open sets instead of just one. Let be the set of points which are not Lebesgue points of , together with the endpoints . This is a null set. Let , and then let be small enough that whenever is measurable with . We can also ensure that .

For every natural number we can find an open set containing of measure . In particular we see that and thus .

Now define a gauge function as follows.

- If , we define to be small enough that the open interval lies in , where is the first natural number such that , and also small enough that holds whenever . (Here we crucially use the
*everywhere*differentiability to ensure that exists and is finite here.) - If , we let be small enough that holds whenever , and such that whenever is an interval containing of length at most , exactly as in the proof of Theorem 89.

Applying Cousin’s theorem, we can find a partition with , together with real numbers for each and .

As before, we express as a telescoping series

For the contributions of those with , we argue exactly as in the proof of Theorem 89 to conclude eventually that

where is the union of all the with . Since

we thus have

Now we turn to those with . By construction, we have

fir these intervals, and so

Next, for each we have and for some natural number , by construction. By countable additivity, we conclude that

Putting all this together, we again have

Since was arbitrary, the claim follows.

Remark 93The above proposition is yet another illustration of how the property of everywhere differentiability is significantly better than that of almost everywhere differentiability. In practice, though, the above proposition is not as useful as one might initially think, because there are very few methods that establish the everywhere differentiability of a function that do not also establish continuous differentiability (or at least Riemann integrability of the derivative), at which point one could just use Theorem 11 instead.

Exercise 94Let be the function defined by setting when is non-zero, and . Show that is everywhere differentiable, but the deriative is not absolutely integrable, and so the second fundamental theorem of calculus does not apply in this case (at least if we interpret using the absolutely convergent Lebesgue integral). See however the next exercise.

Exercise 95 (Henstock-Kurzweil integral)Let be a compact interval of positive length. We say that a function is Henstock-Kurzweil integrable with integral if for every there exists a gauge function such that one haswhenever and and are such that and for every . When this occurs, we call the

Henstock-Kurzweil integralof and write it as .

- Show that if a function is Henstock-Kurzweil integrable, it has a unique Henstock-Kurzweil integral. (
Hint:use Cousin’s theorem.)- Show that if a function is Riemann integrable, then it is Henstock-Kurzweil integrable, and the Henstock-Kurzweil integral is equal to the Riemann integral .
- Show that if a function is everywhere defined, everywhere finite, and is absolutely integrable, then it is Henstock-Kurzweil integrable, and the Henstock-Kurzweil integral is equal to the Lebesgue integral . (
Hint:this is a variant of the proof of Theorem 89 or Proposition 92.)- Show that if is everywhere differentiable, then is Henstock-Kurzweil integrable, and the Henstock-Kurzweil integral is equal to . (
Hint:this is a variant of the proof of Theorem 89 or Proposition 92.)- Explain why the above results give an alternate proof of Exercise 10 and of Proposition 92.

Remark 96As the above exercise indicates, the Henstock-Kurzweil integral (also known as theDenjoy integralorPerron integral) extends the Riemann integral and the absolutely convergent Lebesgue integral, at least as long as one restricts attention to functions that are defined and are finite everywhere (in contrast to the Lebesgue integral, which is willing to tolerate functions being infinite or undefined so long as this only occurs on a null set). It is the notion of integration that is most naturally associated with the fundamental theorem of calculus for everywhere differentiable functions, as seen in part 4 of the above exercise; it can also be used as a unified framework for all the proofs in this section that invoked Cousin’s theorem. The Henstock-Kurzweil integral can also integrate some (highly oscillatory) functions that the Lebesgue integral cannot, such as the derivative of the function appearing in Exercise 94. This is analogous to how conditional summation can sum conditionally convergent series , even if they are not absolutely integrable. However, much as conditional summation is not always well-behaved with respect to rearrangement, the Henstock-Kurzweil integral does not always react well to changes of variable; also, due to its reliance on the order structure of the real line , it is difficult to extend the Henstock-Kurzweil integral to more general spaces, such as the Euclidean space , or to abstract measure spaces.

## 160 comments

Comments feed for this article

17 October, 2010 at 6:13 am

MathStudentDear Prof. Tao,

In a recent lecture post on generalizing Lebesgue integrals (sorry I don’t know how to link), you mention that it is usually not useful to define “improper” Lebesgue integrals or the analogs of Cauchy Principal Value type integrals. Are there other types of integrals that work in those cases? With possibly, some additional restrictions on the functions being integrated?

Thanks.

17 October, 2010 at 11:42 am

Terence TaoYes, at least in one dimension; see for instance the discussion of the Henstock-Kurzweil integral in Exercise 52 and Remark 18. This integral can handle improper integrands which are only conditionally convergent rather than absolutely convergent around a finite singularity. However, it does not cover principal value integrals around a singularity. For these, complex-analytic methods, such as analytic continuation, can sometimes be used to evaluate these integrals, although these are quite different techniques than the measure-theoretic ones presented here.

17 October, 2010 at 7:01 am

Real Analysis (=Measure Theory) by Terence Tao « UGroh's Weblog[…] Lecture 2: Das Lebesguesche Integral Lecture 3: Maße und Maßräume Lecture 4: Konvergenzbegriffe Lecture 4: Differentiation […]

17 October, 2010 at 8:16 am

Wang XiuliIt is not a comment.So long is the post,that I think it must be part of a lecture for your class.

21 October, 2010 at 12:46 pm

Marco AnguloThere is a typo at the end of the proof for Theorem 4. The expression \int_0^1 f(x) dx should be replaced by \int_0^1 f(x) dt.

[Corrected, thanks – T.]23 October, 2010 at 2:27 am

AnonymousThe last sentence in Ex. 21 is incomplete

[Corrected, thanks – T.]24 October, 2010 at 1:05 pm

AnonymousDear Prof. Tao,

Could you please add some notes on Taylor’s expansion thms to this lecture notes?They are always used in Mathematics and it would be nice to review here.

thanks

25 October, 2010 at 10:44 am

Constantin NiculescuIn Exercise 41, the inequality defining a convex function should be reversed.

[Corrected, thanks – T.]30 October, 2010 at 6:54 pm

245A, Notes 6: Outer measures, pre-measures, and product measures « What’s new[…] is a jump function (as defined in Definition 17 of Notes 5), show that is equal to the linear combination of delta functions (as defined in Exercise 22 of […]

3 November, 2010 at 1:18 am

AnonymousThe dominator of \theta_x (fraction) defined in the proof of Lemma. 18, 3. should be “F_+(x) – F_-(x)” instead of “F_+(x) – F(x)”.

[Corrected, thanks – T.]3 November, 2010 at 6:51 pm

Asymptotic decay for a one-dimensional nonlinear wave equation « What’s new[…] differentiable. (By coincidence, I am teaching this theorem in my current course, both in one dimension (which is the case of interest here) and in higher dimensions.) A compactness argument allows one […]

5 November, 2010 at 11:52 am

chandrasekharRespected Sir,

There are some theorem’s measure theory which guarantee “Almost everywhere convergence” is enough for showing everywhere convergence.

Similarly are there any theorems which guarantee that almost everywhere differentiability is enough for showing everywhere differentiability.

9 December, 2010 at 2:42 pm

J.P. McCarthyI have posted a theorem here below which might be relevant to your question and indeed you may be able to help me.

J.P.

8 November, 2010 at 2:42 am

cyEx.47#5. uniformly continuous is repeated twice, i suppose the first should be just continuous.

[Corrected, thanks – T.]17 November, 2010 at 5:08 pm

YasserDear Prof. Tao,

In the proof of the rising sun lemma did you mean in the definition of the set A?

Thanks

17 November, 2010 at 7:45 pm

Terence TaoEither definition would work in this case. (Because lies outside , no element of can exceed anyway.)

23 August, 2017 at 1:49 am

Joe LiBut later I don’t see why . However I agree that

[Explanation added as to why the elements of do not reach or exceed . -T]24 August, 2017 at 2:32 am

Joe LiThank you Prof. Tao.

But what I referred to is that the line below: . is the right end point of the whole interval, and there is no reason that is covered by , for example when is decreasing in .

[Inclusion corrected to . -T]21 November, 2010 at 4:16 pm

AnonymousHi, Prof. Tao

I have two questions about exercise 26:

1, why P is radially non-increasing but P~ is a non-decreasing function?

2, in the question 1, what do you mean by x^2, because x is d-dimension vector. Do you mean by the inner product?

Thank you.

[Oops, these were typos. Corrected, thanks – T.]22 November, 2010 at 11:49 pm

LeiSeveral typos:

1, Exercise 18: not just in 2B’_j, it should be 3B’_j

2, Exercise 21: the left hand side, du(y) should be du.

3, Exercise 41: f is convex, so its derivative f’ should equal a.e. to a monotone non-decreasing function, not non-increasing.

[Corrected, thanks – T.]25 November, 2010 at 2:03 pm

LeiIn question 3 of exercise 46, we can further prove that |F_{n+1}|-F_{n}|=<2^(-n-1), because it is easy to verify that |F_1-F_0|=<1/2 and we can induct on n.

25 November, 2010 at 4:57 pm

XttIn Exercise 26, is it that P is actually radially non-increasing? Thanks.

[Corrected, thanks – T.]25 November, 2010 at 6:59 pm

ผลบอลThe last sentence in Ex. 21 is incomplete . Great!

[I don’t see the issue here – could you clarify? – T.]26 November, 2010 at 9:50 pm

AnonymousProf Tao,

Should F(x) in Exercise 47.6 be called indefinite integral or definite integral?

29 November, 2010 at 9:58 am

dukezhangIt seems the example following Definition 17 is not non-decreasing. Since J(q_1)=1, J(q_2)=.5, but q_1<q_2. Is that the case?

[Oops, the jumps were going the wrong way; this has been fixed. Incidentally, the are not (and cannot) be ordered in a monotone fashion, as the rationals are not order-equivalent to the natural numbers. – T.]29 November, 2010 at 9:50 pm

HildaIn Ex.20, should it be “integration of |f|” instead of “integration of f”?

[Corrected, thanks – T.]29 November, 2010 at 9:54 pm

HildaSorry, it’s Ex. 21.

30 November, 2010 at 4:50 pm

FarzinDear Professor Tao,

In the 4th, 3th and 2nd to the last equation before Exercise 49, should some of the sum notations be integrals?

[Corrected, thanks – T.]2 December, 2010 at 2:08 pm

Bunyamin SariIn Weierstrass function, exercise 27, does the coefficient 2 in Sine function really work? If a straightforward estimate to be used, shouldn’t it be bigger than 4?

[Corrected, thanks – T.]6 December, 2010 at 4:57 pm

AnonymousDear Prof. Tao,

I believe the following is a counter example to 50: Take and to be any positive, continuously differentiable bump function that is compactly supported in (a, b). Then .

[Yes, but in this caseand so.– T.]6 December, 2010 at 5:41 pm

AnonymousDear Prof. Tao,

In the second paragraph of the proof of proposition 25, it states that

$\int_{\bigcup_{m=1}^\infty U_m} |F'(x)|dx \leq \epsilon$.

I don’t believe this is true, unless the open sets $U_m$ are nested and $U_1 = U$. This wasn’t used any where else in the proof, though.

Some clarifications added – T.]6 December, 2010 at 9:59 pm

AnonymousThanks – got it!

9 December, 2010 at 2:40 pm

J.P. McCarthyStudents of Prof. Tao (or even the man himself even better!),

Could you please tell me have I made a mistake in this proof. I believe the theorem is correct but after struggling with finding a proof I think the following seems a bit too good to be true:

Theorem

Suppose that is differentiable except perhaps at and suppose that is continuous at , . is continuous on and differentiable on . Then, by the Mean Value Theorem, such that

Now taking the limit as on both sides:

Now as hence the left hand side is equal to :

Similarly taking and we can show:

Hence is differentiable at with derivative

Full details at http://irishjip.wordpress.com/2010/12/09/ms-2001-differentiability-question/

9 December, 2010 at 2:48 pm

J.P. McCarthyARRGGGHHHH!! Please just follow the link, the proper statement of the theorem is at the end.

Cheers,

J.P.

8 January, 2011 at 11:53 am

AnonymousDear professor,

In step 1 of lemma 13’s algorithm, the expression “(which, initially, will be all the balls )” is a bit ambiguous. It is not immediately clear whether these balls are the or the balls . You may want to alter that.

Thank you for the attention!

[Reworded, thanks – T.]11 January, 2011 at 8:36 am

AnonymousDear professor,

In lemma 16, last time you wrote , did you mean ?

Thank you!

[Corrected, thanks – T.]13 May, 2011 at 7:49 am

Stein’s maximal principle « What’s new[…] to pointwise almost everywhere for all in , and not just in the dense subclass. See for instance these lecture notes of mine, in which this method is used to deduce the Lebesgue differentiation theorem from the […]

31 October, 2012 at 4:42 pm

AnonymousWhat do the notations F’\phi(x) and F\phi'(x) mean in Exec 51? I cannot find the definitions.

31 October, 2012 at 8:52 pm

Terence Taois the derivative of , multiplied by , and similarly for .

29 November, 2012 at 3:27 pm

jdemDear Tao, If I remember correctly, the Cantor function is differentiable almost everywhere in the Cantor set (ex 47.8), and in this kind of selfsimilar sets with Hausdorff dimension d, the points where the Devil’ staircase is not differentiable has dimension d^2 < d.

22 March, 2013 at 8:41 am

JackProf. Tao, what’s the difference between Theorem 4 and Corollary 5? Isn’t the later one just part of Theorem 4 since ?

22 March, 2013 at 9:50 am

Terence TaoYes; this is why the latter result is labeled as a corollary of the former one.

5 December, 2013 at 10:48 am

shubhashedthikereProf. Tao,

I want to know why in the definition of functions with ‘bounded variation’, the supremum is taken over “finite” increasing sequences, why “countable” increasing sequences are not allowed?

5 December, 2013 at 10:50 am

Terence TaoThese would give equivalent definitions (the two suprema are equal, thanks to (an easy case of) the monotone convergence theorem).

18 July, 2014 at 9:53 pm

Stein’s spherical maximal theorem | What's new[…] for all and almost every . See for instance my lecture notes on this topic. […]

16 November, 2014 at 6:35 pm

AnonymousIn Exercise 1, do you actually mean “ need not be continuous” in the last line?

17 November, 2014 at 6:54 am

AnonymousAh, I didn’t see Exercise 2. A simple function like where is a proper sub-interval is an example in this case.

17 November, 2014 at 7:32 pm

JackThis argument requires two ingredients, which we state informally as follows:A verification of the convergence result for some “dense subclass” of “nice” functions {f}, such as continuous functions, smooth functions, simple functions, etc.. By “dense”, we mean that a general function {f} in the original class can be approximated to arbitrary accuracy in a suitable sense by a function in the nice subclass.A quantitative estimate that upper bounds the maximal fluctuation of the linear expressions {T_h f} in terms of the “size” of the function {f} (where the precise definition of “size” depends on the nature of the approximation in the first ingredient).

I’m trying to identify the second ingredient of the density argument in the proof of Theorem 7. As I see from the proof where Lemma 9 is used, “fluctuation” is bounded by . Is this the “’size’ of the function “? (It doesn’t make sense though, I don’t know what would be the size here.)

18 November, 2014 at 10:22 am

Terence TaoLemma 9 bounds the weak norm of maximal value of in terms of the norm of . So in this case, the relevant size of is the norm.

19 November, 2014 at 1:10 pm

JackAh, so the second step in the density argument is to bound instead of and in the example you give in the proof of Proposition 8, should be understood as ? (Still a little bit confused about what one should bound in the “quantitative estimate” step.)

I didn’t find the definition until I skimmed through the 245C note. In Chapter 3 of Stein-Shakarchi, the inequality in Lemma 9 is called to be of “weak type.” I didn’t understand why it is called in that way. (I guess it is because the “weak norm” as you put is used. )

The quantity on the left hand side of the inequality looks very much like the distribution function in probability. If one replace the Lebesgue measure with some probability measure, the inequality still holds?

19 November, 2014 at 12:44 pm

AnonymousI have no intuition at all why Cousin’s Theorem can be true and what might be a possible motivation for such statement. (Unlike the rising sum lemma for instance, one can vividly sense why the statement should be true). It looks like an indispensable tool for define the Henstock–Kurzweil integrals. But how should one understand the “gauge function” (and the whole statement)? (It looks something like the “norm” of a partition when we define Riemann integral though.)

30 November, 2014 at 8:19 pm

Alperen SirinMathematical Proofs! Yes! I’ve been looking for this for so long now! Well, although I found a lot of interesting websites and articles about the background proof of calculus, it is quite pleasing to find it all together.

30 November, 2014 at 8:54 pm

Alperen SirinReblogged this on Ideas, Thoughts, and Research and commented:

Differentiation Theorems from Terence Tao

21 December, 2014 at 6:29 pm

AnonymousCan one do Exercise 5 without essentially using the property that the measure defined as is absolutely continuous w.r.t. the Lebesgue measure?

[One can also use dominated convergence – T.]21 December, 2014 at 7:13 pm

AnonymousTypo in the proof of Proposition 8, “to the absolutely integrable function “?

[I don’t see the issue here – T.]22 December, 2014 at 5:57 am

AnonymousIn the proof of Proposition 8:

… Applying Littlewood’s second principle (Theorem 15 from Notes 2) to the absolutely integrable function , we can find a continuous, compactly supported function…

I think you mean “the absolutely integrable function …”

[Corrected, thanks – T.]21 December, 2014 at 7:55 pm

Anonymoustypo right after formula (6), “whenever is sufficiently close to ”

[I don’t see the issue here – T.]22 December, 2014 at 6:02 am

Anonymous“Now let . From the dense subclass result (Corollary 5) applied to the continuous function , we have

whenever is sufficiently close to . Combining…”

Should it be “h is sufficiently close to 0”?

[Corrected, thanks – T.]23 December, 2014 at 6:50 am

Anonymous…we conclude that

for almost every . If we then let go to zero along a countable sequence (e.g. for , we conclude that

for almost every , and the claim follows.

———————————————

Why would one need to pass the to a sequence? If one has a nonnegative quantity such that $X0$, then . Is it just a matter of writing style or is there something wrong with my thoughts? (This might be a stupid question. I ask it just because you’ve mentioned the “-trick” over and over again in the previous notes and there seems no need to use the sequence here…)

[The need to pass to a countable sequence is to ensure that the total exceptional set of ‘s remains of measure zero. -T.]23 December, 2014 at 1:46 pm

AnonymousAh, for different , the “exceptional set” of ‘s might be different. Fair enough. Could have been clearer if one denotes the exceptional set as in the beginning of the proof(-:

31 December, 2014 at 1:27 pm

AnonymousIs “ being of bounded variation on ” also a

necessarycondition for the existence of ?3 January, 2015 at 8:57 am

Terence TaoDepends on what one means by here, but with the classical definition, the answer is no; consider for instance the function .

4 January, 2015 at 3:44 pm

AnonymousIs the absolute continuity of necessary for

to be true where is with the classical definition?

Does the negative answer in your last comment suggest that we might have another class of functions so that we still have the identity (or the inequality in the Prop. 22) with understood as in the classical sense?

4 January, 2015 at 5:11 pm

Terence TaoIf is required to be absolutely integrable, then yes; see part 6 of Exercise 48.

If is not required to be absolutely integrable, then one has to decide how one is to interpret the integral , as one can no longer use the Lebesgue integral. There are other integration concepts for which one can broaden the range of the second fundamental theorem of calculus, such as the Henstock-Kurzweil integral; see Exercise 53.

Incidentally, these sorts of questions are excellent exercises for you to work out on your own (cf. https://terrytao.wordpress.com/career-advice/ask-yourself-dumb-questions-%E2%80%93-and-answer-them/ ).

7 January, 2015 at 7:01 pm

AnonymousWhat could be wrong if is unbounded in Lemma 18.3?

What I understand from the proof is that one could have and thus by

definition, is not a jump function. But I don’t have a concrete counterexample that Lemma 18.3 could be false if is unbounded.What’s more, is there any significant disadvantage if we allow just the point-wise limit of in the definition of jump functions?

[See Exercise 33 – T.]27 November, 2015 at 1:46 pm

JohnProfessor Tao:

Could you please explain to me why, in the Rising Sun Lemma, the second condition is fulfilled even when taking x=a? It seems to me that if one takes then is not in the union of the and yet for each .

[Corrected, thanks – T.]18 December, 2015 at 12:31 pm

JohnProfessor Tao:

A proof of Cousin’s Theorem seems straightforward enough with a bisection type argument, however I haven’t been able to follow your suggestion of using the Besicovitch covering lemma together with the Heine-Borel Theorem. Could you please give some additional hint on this approach?

Thanks.

[Apply the Heine-Borel theorem to the intervals , followed by the Besicovitch covering lemma. I’d be curious to see your bisection argument. -T.]18 December, 2015 at 1:34 pm

JohnI came up with the following two arguments:

Proof 1. Notice that if such a tagged partition exists in the intervals and by piecing together the two partitions we may construct one that works for the entire interval . The argument proceeds by contradiction; suppose the partition does not exist in .Then bisect ; choose a half in which a suitable partition does not exist (it cannot exist in both halves). Keep doing this to find a shrinking sequence of nested closed intervals in each of which we may not find a suitable partition. Their intersection is non-empty; take in said intersection, then by taking some shorter than we see that the trivial partition will work in this interval, by tagging the point $t$; a contradiction.

Proof 2. Consider . By the similar piecing-together arguments as in Proof 1, show that and then show that .

Do these seem correct?

Your suggestion is precisely what I had already attempted but I don’t see how the finite covering given by the lemma may be used to obtain a suitable partition (the naive approach of simply taking the partition to be all endpoints of the intervals doesn’t seem to work).

18 December, 2015 at 1:44 pm

Terence TaoThese arguments look correct to me (and are closely related to either the Bolzano-Weierstrass theorem or the Heine-Borel theorem, which they should as all these results involve the compactness of .)

I should have used the intervals in place of . By Heine-Borel and Besicovitch one can find intervals covering such that any point lies in at most two of the intervals. One can delete any interval which is fully contained in any other interval. If one then lines up the in increasing order, then the intervals and must intersect. Now pick to be a common point of these intervals (with appropriate conventions at the endpoints ).

18 December, 2015 at 1:51 pm

JohnI see what you had in mind now, thank you very much.

3 March, 2016 at 12:22 pm

AnonymousLet and be bounded open and connected. Consider a function with . Speaking of differentiation of with respect to the variable , it seems that there are two ways to look at it: (1) the

partialderivative (2)thederivativeof , where is regarded as a function from to some function space which is a subset of . Are these two interpretations the same when one talks about differentiation?3 March, 2016 at 2:09 pm

Terence TaoPretty much yes, assuming either that has sufficient regularity and decay, or else that the derivatives are being interpreted in an appropriately weak sense (e.g. distributional derivative, or using a weak topology for the function space).

4 March, 2016 at 7:01 am

AnonymousIt is easy to state Definition 19, but I’m really wondering if one could apply it to check whether a specific function is of bounded variation or not. (Why is it a useful concept then?) It looks pretty much an “integral definition”. Does it relate to the line integrals in calculus?

4 March, 2016 at 8:28 am

Terence TaoOne can use Exercise 38 or Proposition 20 as ways to verify bounded variation in practice. Bounded variation implies several additional useful properties, such as almost everywhere differentiability (Corollary 21); for functions of bounded variation on a half-line, it also implies existence of a limit at infinity, which is useful for instance in ergodic theory.

4 March, 2016 at 7:27 am

AnonymousFor Remark 2, if is differentiable in and continuous in , must it be differentiable in also?

4 March, 2016 at 8:26 am

Terence TaoNo; the standard counterexample is on (with ).

4 March, 2016 at 11:56 am

AnonymousIn Theorem 24, do we have that the converse is actually also true, namely, if the identity

makes sense, then must be absolutely continuous on ?

If is differentiable, can we conclude that is absolutely continuous on so that Proposition 25 is implied by Theorem 24?

4 March, 2016 at 3:41 pm

Terence TaoTheorem 6 serves as a converse to Theorem 24. Note that just having well-defined and absolutely integrable and for a single choice of is not sufficient (take for instance to be the Cantor function from Exercise 47, but with set equal to 0 rather than 1).

Incidentally, all of these questions are great exercises to work out yourself; see https://terrytao.wordpress.com/career-advice/ask-yourself-dumb-questions-%E2%80%93-and-answer-them/

4 March, 2016 at 8:48 pm

AnonymousSuppose is differentiable. It is true that

but not necessarily true that

for all , which implies that is absolutely continuous.

I guess this is your point. But I don’t have a counterexample for the “not necessarily true” part. Do you have a hint how to find it?

5 March, 2016 at 7:22 am

AnonymousWell, is not even uniformly continuous on . And thus it cannot be absolutely continuous. But it is .

11 April, 2016 at 9:58 am

Fonction maximale de Hardy-Littlewood, théorème de différentiation de Lebesgue | Matthieu Joseph[…] constante n’est pas optimale dans cette inégalité. Terence Tao propose un exercice (l’exercice 19) pour améliorer par , puis il discute ensuite de la […]

4 May, 2016 at 2:09 pm

AnonymousIn Exercise 7, is measurable on ?

4 May, 2016 at 2:33 pm

AnonymousIn Exercise 7 should be well defined for a.e. instead of every ?

5 May, 2016 at 8:18 am

Terence TaoIn this case can be shown to be well defined for all (the integrand is absolutely integrable).

Also, the measurability of and (and hence of the product) is easily deduced from observing that the composition of a measurable function and a continuous map (or more generally, a measurable map) is again measurable. Here, the maps in question are and .

5 May, 2016 at 4:58 pm

AnonymousIf is Borel measurable, then is Lebesgue measurable or Borel measurable whenever is. But we only know that is Lebesgue measurable. How do you conclude that is (Lebesgue) measurable where ?

5 May, 2016 at 5:05 pm

AnonymousContinuity of itself is not enough in this issue I suppose? It seems that one need to show directly that is Lebesgue measurable whenever is.

6 May, 2016 at 8:19 am

Terence TaoFair enough; the map defined by is continuous but the preimage of the null set is nonmeasurable for any nonmeasurable . But in this particular case one can check that the preimage of a set of zero outer measure is again of zero outer measure, so null sets pull back to null sets. Since Lebesgue measurable sets differ by a null set from a Borel measurable set, the claim for Lebesgue sets then follows from the corresponding claim for Borel sets.

One can also use Exercise 21, 22 of Notes 1.

9 May, 2016 at 1:49 pm

AnonymousIn Stein and Shakarchi, is assumed to be integrable on (so is ) and the conclusion is that is well defined for a.e. . Do you have a hint that why one cannot have being well defined for all in that case? Is there a counterexample showing what could go wrong?

9 May, 2016 at 2:40 pm

Terence TaoConsider for instance . The Cauchy-Schwarz inequality tells us that this integral is well-defined when are square-integrable, but when are merely integrable then this integral may be divergent (e.g. consider the case and ).

9 May, 2016 at 6:49 pm

Anonymous2I read through notes 1 up to this note but I can’t find a useful statement to show that is well-defined for a.e. if one only assumes that are merely integrable… The comment above seems to suggest that this is unlikely true?

9 May, 2016 at 9:42 pm

Terence TaoIf are absolutely integrable, then by the Fubini-Tonelli theorem, is absolutely integrable on , which by further application of Fubini-Tonelli shows that is absolutely integrable in for almost every .

7 May, 2016 at 6:19 am

Anonymous“But in this particular case one can check that the preimage of a set of zero outer measure is again of zero outer measure”

Would you elaborate how this is done? (What’s more, why did you use “outer measure” instead of “Lebesgue measure”?)

(Trying to prove by contradiction seems also difficult: Let with . Suppose is a subset of with positive Lebesgue measure. How can I show that can not have measure zero?)

7 May, 2016 at 7:53 am

Terence TaoOne needs to use outer measure initially because one does not know

a priorithat the preimage is measurable.A warmup question would be to show that if has zero outer measure, then the set has zero outer measure; using the countable subadditivity of outer measure, this implies that also has outer measure. This gives the claim for ; the case of is similar.

As I said in my previous post, these claims are special cases of Exercises 21 and 22 of Notes 1.

12 May, 2016 at 6:32 am

AnonymousRegarding the concept “good kernels” in Exercise 27, Stein and Shakarchi give two sets of conditions.

One is

(1)

(2)

(3) For every , as

Another one is

(1)

(2′)

(3′) for all and .

You also give a set of condition in Exercise 27.

(1)and has total mass equal to .

(2″)non-negative,

(3″) radial,

(4″) radially non-increasing,

Other than the “total mass equals to 1”, are other conditions in each set related to each other? In practice, how does one choose which set of conditions to use?

12 May, 2016 at 8:32 am

Terence TaoBasically it depends on the application. In analysis there often is not a “one-size-fits-all” definition for a given concept (such as a “good kernel”), one often has to tailor the precise axioms for the need at hand, strengthening them if this will help prove other theorems, or weakening them if this helps find instances of the concept. For the purpose of an introductory course, one often does not strive for the most general statement one can make in these directions, but instead focuses on a simple illustrative case, even if this means that the axioms are unnecessarily strong or the conclusion unnecessarily weak.

This is in contrast with the more algebraic areas of mathematics, in which there often are natural and canonical definitions for key concepts; cf. Gowers’ “Two cultures of mathematics” essay.

23 May, 2016 at 12:15 pm

ChrisDear Terry,

in the proof of Lemma 18.2 you use this nice map. You choose for each jump at a rational number which lies between the left and right limit. Do you use the axiom of choice to do this?

Thanks.

23 May, 2016 at 6:37 pm

Terence TaoOne does not need the axiom of choice for this, as it is possible to place an explicit well-ordering on the rationals which can be used to create a choice function.

19 July, 2016 at 6:32 pm

Suntingdear prof. tao. i am wondering whether the rising sun lemma could be true when it is defined on R. i.e.,

for any continuous F:R->R;

then we could find at most countable famility of disjoint non-empty open intervals.In=(an,bn);such that:

1)for each n, F(an)=F(bn);

2)for any x not belonging to any of In. then we have F(y)<=F(x)for any x<=y;

19 July, 2016 at 6:57 pm

Suntingsorry , i may be wrong . it seems this lemma dont admit In=(-inf,a) or In=(a,inf);

27 September, 2016 at 10:52 am

246A, Notes 2: complex integration | What's new[…] variants of the above integrals (e.g. the Henstock-Kurzweil integral, discussed for instance in this previous post), which can handle slightly different classes of functions and have slightly different properties […]

2 December, 2016 at 4:58 pm

AnonymousIn Theorem 6, do we have more information regarding the such that . For instance, do we a characterization of such ? Suppose is continuous at . Can we conclude that ?

3 December, 2016 at 2:19 am

Anonymousholds whenever is a Lebesgue point of (see e.g. the Wikipedia articles on Lebesgue points and Lebesgue differentiation theorem). Note that each continuity point of is a Lebesgue point of .

22 December, 2016 at 8:01 am

coupon_clipperTerry, in exercise 9 part 1, you say to show that has positive measure for some z. I get the feeling that this is supposed to be simple but I’m stuck here.

I was able to show that it this weren’t the case, then there would be a null set (in ) that maps to the entire complex plane, but I’m not sure that leads to a contradiction. Can you give a hint?

[If is null for every , then one should be able to show that all of is null by expressing the latter set as a countable union of sets of the former type. -T.]23 December, 2016 at 5:05 am

coupon_clipperThat did it. Thanks T!

25 December, 2016 at 8:50 am

AnonymousThere are so many inequalities you labeled as “Hardy-Littlewood”. Did they actually give all of them mentioned here?

Besides, in “Just as the integration theory of unsigned functions can be used to develop the integration theory of the absolutely

convergentfunctions”, I think you mean “absolutely integrable functions”?26 December, 2016 at 10:35 am

Terence TaoThanks for the correction. All the one-dimensional Hardy-Littlewood inequalities stated here are essentially in the original 1931 paper of Hardy and Littlewood, although the notation is somewhat different. The higher-dimensional generalisations of the Hardy-Littlewood inequality first appear I believe in a 1939 paper of Wiener, so strictly speaking it should be referred to as a maximal inequality of Hardy-Littlewood type.

25 December, 2016 at 9:03 am

AnonymousIs the converse of Corollary 21 also true so that we have a characterization of differentiable almost every where functions?

26 December, 2016 at 10:47 am

Terence TaoNo, for instance is differentiable almost everywhere on the real line (after assigning some arbitrary value to ) but has infinite variation; another class of examples would be functions that are locally constant outside of a Cantor set, but whose values on each connected component of the complement of the Cantor set are set to arbitrary values unrelated to each other. I would imagine that there is no particularly useful characterisation of the entire class of almost everywhere differentiable functions other than the tautological one of being differentiable outside of a set of measure zero.

25 December, 2016 at 6:10 pm

AnonymousWhat is the relation between bounded variation and integrability on a compact interval?

[The fundamental theorems of calculus identify the absolutely continuous functions as the antiderivatives of the integrable functions, and the absolutely continuous functions are a subclass of the bounded variation functions. -T.]26 December, 2016 at 1:15 pm

AnonymousSorry for the confusion. I was going to ask what is the relation between bounded variation and integrability

of a functionon a compact interval.Say . Is there any relation between the integrability (being absolutely integrable or not) of on $[a,b]$ and the “bounded variationess” of ?

26 December, 2016 at 4:45 pm

AnonymousA “d” is missing in the statement of Theorem 12.

[Corrected, thanks – T.]26 December, 2016 at 8:35 pm

Terence TaoFunctions of bounded variation are bounded and measurable, hence integrable. On the other hand, the converse statements are very far from being true; functions of bounded variation have roughly one degree of regularity, functions that are merely integrable have none.

3 January, 2017 at 4:21 pm

AnonymousAll the dumb questions I asked here are for doing the exercise in Stein-Shakarchi:

Let and $f:[0,1]\to\mathbb{R}$ be defined with for and . Show that is BV if and only if $a>b$.

When , all I can find is

$f’\in L^1([0,1])$. I really don’t know if I can conclude that is BV due to the possible bad behavior at since does not necessarily have a derivative there.

1 January, 2017 at 9:00 am

AnonymousThis is a related “dumb question” that I don’t know how to answer. Consider . (1)Suppose is BV on . Then one has is differentiable a.e.. Can one say further that ? (2) One the other hand, if (assuming exists a.e.), can one conclude that is BV on $[a,b]$?

I can only answer this question when is absolutely continuous, which can be done by Exercise 38.

[For (1), combine Proposition 20 with Proposition 22. for (2), see my comment from 26 Dec 10:47am. -T.]2 January, 2017 at 11:44 am

AnonymousThanks for the comment. For (2) I think the example on does not work since the derivative is not on and you are referring to the Cantor-function-like function actually?

One can at least now say that (i)being differentiable a.e. on and (ii)the derivative are

necessaryfor being BV on but far less thansufficientdue to the existence of counterexamples (the Cantor-function-like function?).What if one makes the exceptional set for (i) smaller: is differentiable everywhere on except at one point (or finitely many points even countably many)?

3 January, 2017 at 3:36 pm

AnonymousI’m trying to use Exercise 36. Does one have

3 January, 2017 at 3:44 pm

AnonymousAnd of course one should also assume that is continuous.

26 December, 2016 at 12:41 pm

AnonymousIn exercise 28, Weierstrass function is represented by a lacunary Fourier series. Is it necessary for any such example (periodic, continuous and non-differentiable everywhere) to have a lacunary Fourier series representation?

26 December, 2016 at 8:37 pm

Terence TaoNo. For instance, one can show that non-differentiable everywhere periodic functions are a comeager subset of the class of continuous periodic functions, whereas the lacunary Fourier series are a meager set.

26 December, 2016 at 6:03 pm

AnonymousDue to my ignorance, I found the proof of the differentiation theorem (between Lemma 9 and Lemma10) hard to follow while the counterpart in Stein-Shakarchi (p105) seems to be very clear. (I like your comment about the “density argument” a lot. But I feel very stupid and frustrated that I didn’t figure out how the density argument is made.) Am I missing something that makes the difficulties of reading this particular proof? It seems that the style of writing here is sort of improvising, which is rather different from your advice on writing(https://terrytao.wordpress.com/advice-on-writing-papers/).

28 December, 2016 at 12:25 pm

AnonymousI stared at the definition of Dini derivatives and wanted to try some nontrivial examples but I couldn’t figure out a way. Let for and . Would you give an example for how to calculate one of the four Dini derivatives?

In general, is there a systematic way to calculate the Dini derivatives at the point where the given function is not differentiable?

28 December, 2016 at 1:28 pm

AnonymousWould you elaborate the hint for Exercise 30? What does it mean by ranges over a countable set and how it would be useful?

[Countable unions or intersections of measurable sets stay measurable, whereas uncountable unions or intersections do not. A naive description of the level set of a Dini derivative, e.g. , will involve an uncountable union or intersection because ranges over the set of positive reals, which is an uncountable set. If one can somehow restrict the range of to a countable set (e.g. the set of positive rationals), then this problem goes away. -T.]12 January, 2017 at 5:56 am

coupon_clipperExercise 28 (proving the Weierstrass function is nowhere differentiable — the continuous part was easy for me) is really giving me trouble. Following your hint, I computed and (I switched to using k as an index so we’re not using n twice) and found that they can both be expressed as a sum of terms instead of an infinite series.

Then when I subtract them, I can use the “difference of sines” formula, but that’s where I get stuck:

I don’t see any way to bound this from below (in absolute value).

I know this seems like a basic analysis problem and it wasn’t really the point of this whole chapter, but I really want to figure it out.

12 January, 2017 at 9:23 am

Terence TaoHmm, actually the hint would fit the problem better if the exercise used cosines rather than sines, so that there is an term that easily dominates the rest of the sum. I’ll change the exercise accordingly.

12 January, 2017 at 11:13 am

coupon_clipperThanks, Terry! That’s better. (You swapped an m and n in part 2 though.)

[Corrected, thanks – T.]12 January, 2017 at 11:14 am

coupon_clipperAlso, it’s exercise 52 in case anyone else is reading along. I’m not sure where I got 28 from.

1 May, 2017 at 2:32 pm

PierreThanks Tao for this lectures.

Maybe i don’t understand the formulation of the Besicovitch covering lemma :

If i take (0,1),(-1,2/3),(1/3,2), there is no subfamily with the same union and 1/2 will always be in 3 intervals.

1 May, 2017 at 2:34 pm

PierreAll wrong, sorry

23 August, 2017 at 3:57 am

Joe LiA small typo:

In the proof of Proposition 19:

Applying (4), we conclude that

…

which we rearrange as

the next line there is an extra before the first .

[Corrected, thanks – T.]23 August, 2017 at 4:01 am

Joe LiAt the end of the proof of Lebesgue differentiation theorem (just above lemma 26), is it necessary that go to zero along a COUNTABLE sequence?

[This is because there is a measure zero exceptional set of for each . In order to ensure that the union of these exceptional sets remains of measure zero, one needs to use only countably many . -T]24 August, 2017 at 2:20 am

Joe LiSmall typo in several places in section 2:

should be .

If I’m correct, these places are:

RHS of the line below Theorem 36 (Hardy-Littlewood maximal inequality)

RHS of the line below “It suffices to verify the claim with strict inequality”

RHS of the line below “By inner regularity, it suffices to show that”

RHS of he line below “establish the dyadic Hardy-Littlewood maximal inequality”

[Corrected, thanks – T.]25 August, 2017 at 3:05 am

Joe LiSmall typos:

3rd lines above Remark 60:

should be .

In the proof of Lemma 62:

As discussed previously, is discontinuous only at

should be

As discussed previously, is discontinuous only at

since there is no in the context.

[Corrected, thanks – T.]20 May, 2018 at 8:00 pm

A short proof of the Hardy-Littlewood maximal inequality | George Shakan[…] Incidentally his proof gives the better constant , though this is well known, see for instance exercise 42 in these notes of Tao. One cute related geometric question is can one improve the constant in Vitali’s covering lemma […]

16 September, 2018 at 9:31 am

254A, Notes 1: Local well-posedness of the Navier-Stokes equations | What's new[…] Exercise 2 Relax the hypotheses of continuity on to that of being measurable and bounded on compact intervals. (You will need tools such as the fundamental theorem of calculus for absolutely continuous or Lipschitz functions, covered for instance in this previous set of notes.) […]

18 September, 2018 at 10:44 am

Alan ChangIn Definition 66 (Bounded variation) and a few places afterwards, F(x_{i+1}) should be F(x_{i-1}).

[Corrected, thanks – T.]1 June, 2019 at 7:45 am

AnonymousI don’t quite understand the conclusion of Theorem 11:

Then the Riemann integral is equal to . In particular, we have whenever is continuously differentiable.

Aren’t the following talking about the same things?

– the Riemann integral is equal to

–

1 June, 2019 at 8:40 am

Terence TaoYes, the second claim is a particular case of the first (continuously differentiable functions have a derivative that is Riemann integrable).

1 June, 2019 at 8:11 am

AnonymousIs there a relation between Theorem 89 and Proposition 92? It seems that the hypothesis of Prop 92:

… let be a differentiable function, such that is absolutely integrable.

implies that of Theorem 89, namely is absolutely continuous, doesn’t it? If true, then Proposition 92 seems a bit redundant. (I may misunderstand the purpose of Theorem 89 though.)

1 June, 2019 at 8:43 am

Terence TaoIt is true that an everywhere differentiable function with absolutely integrable derivative is absolutely continuous, but to prove this one has to

useProposition 92 (together with Exercise 87.6). Note that Proposition 92 is a little subtle because it is not true if the function is merely assumed to be differentiable almost everywhere instead of differentiable everywhere, even when the function is continuous, as the example of the Devil’s staircase function illustrates.1 June, 2019 at 10:18 am

AnonymousRegarding Definition 86, is “absolute continuity” a global concept (like “uniform continuity on an interval”) or a local one (like “continuity” at a point)?

In Exercise 87:

Show that every absolutely continuous function is of bounded variation on every compact interval .

Should one understand “absolute continuity” as on or on ?

1 June, 2019 at 2:20 pm

Terence TaoThe property of being absolutely continuous is localisable, in the sense that if a function is absolutely continuous on an interval (or real line) latex I$, then its restriction to a subinterval $J$ will also be absolutely continuous, and conversely if $I$ is partitioned into subintervals and a function is absolutely continuous on each subinterval (and continuous at the endpoints of the subintervals) then it will be absolutely continuous on the entire domain. However, it is not a pointwise property: there is no meaningful notion of a function being “absolutely continuous at a point”.

1 June, 2019 at 10:57 am

AnonymousLemma 62.3 and the Lebesgue decomposition theorem in the linked notes are both a sort of “decomposition” and they both have the concept of “absolute continuity” in the statement.

But in the former decomposition , “absolute continuity” appears implicitly in the assumption while the later one has “absolute continuity” in the conclusion.

Would you elaborate how Lemma 62.3 fits in the Lebesgue decomposition theorem?

1 June, 2019 at 2:22 pm

Terence TaoThe Lebesgue-Stieltjes measure associated to a monotone function need not be absolutely continuous; it can have both absolutely continuous and singular components. The singular component in turn splits into a singular continuous component and a pure point component. The function is the cumulative distribution function of the sum of the absolutely continuous and singular continuous components, and is the cumulative distribution function of the pure point component.

13 October, 2019 at 4:27 am

AnonymousStein said without proofs at the beginning of his singular integral book that it is a “simple” observation that the maximal function satisfies . I searched around the book without finding anything to prove it. I believe it relates to this set of notes. How can one do that?

13 October, 2019 at 5:33 am

AnonymousIf is identically zero, this inequality is clearly false for any

13 October, 2019 at 8:35 am

AnonymousThe assumption on is of course that is not identically zero.

14 October, 2019 at 8:38 am

Terence TaoFor any and one has

for some depending on , as long as is chosen so that is non-zero, which is possible to do for any non-trivial (which is part of the hypotheses in the remark in Stein’s book on pages 5-6, as is the constraint ).

14 October, 2019 at 10:54 am

AnonymousIt seems that it is possible to make dependent only on if the inequality is required to hold only for sufficiently large (where the “sufficiently large” threshold is dependent only on )

17 October, 2019 at 6:59 am

AnonymousThe existence of can be derived by looking at the contrapositive of the statement: if for all , then one must have by the monotone convergence theorem. Is there any “constructive” proof showing what that should at least be?

If one considers the set , then should contain a “large portion” of . If is bounded, then this is trivial.

17 October, 2019 at 7:35 am

Terence TaoPerhaps it is clarifying to look at a simpler discrete analogue of this question. It is tautological that if a sequence is not identically zero, then there exists an such that ; this is a discrete analogue of the assertion that if a function is not identically zero a.e., then there is an such that . However, in both cases there is no bound on the quantity or . One explanation for this is that the key hypothesis of being “not identically zero” is an open condition, rather than a closed condition, and in particular is certainly not a compact condition. So one does not expect to have any uniform bound on conclusions that rely on such an open condition. (Compare for instance the assertion that if a real number is non-zero, then its reciprocal is finite. This assertion is trivial but one has no uniform bound on the magnitude of because the hypothesis of being non-zero is open. If instead we replaced that hypothesis with a closed condition like then one can now obtain a uniform bound.)

So if one somehow upgrades the hypothesis of being not identically zero to something more quantitative (and closed), then there is a chance of getting a usable bound. For instance if one has some lower bound on or that becomes positive when or is large enough then this would bound the quantity or in the previous conclusions.

18 October, 2019 at 1:28 am

AnonymousIs there a precise definition for the concept “open condition” ?

23 October, 2019 at 6:53 am

AnonymousIn Stein-Shakarchi, the maximal function is defined as

I believe this is equivalent to the one given in this set of notes. Is it simply a matter of taste? Or does the flexibility of balls (not necessarily centered at ) in the above definition make any argument related to maximal functions easier in practice?

Stein’s book on singular integrals has the same version of maximal functions as the one here. I am wondering if there is any technical reason that he changed the definition to a different one. (Also a different version of the covering lemma was used there with the constant instead of . I’m curious who in history gave this slight improvement.)

23 October, 2019 at 7:50 am

AnonymousThis is the non-centered version of the maximal function. Hardy-littlewood maximal function is the centered one. The two versions are clearly equivalent since the centered maximal function is upper bounded by the non-centered one which in turn is upper bounded by times the centered one (by doubling the ball radius)

14 May, 2020 at 7:26 am

247B, Notes 4: almost everywhere convergence of Fourier series | What's new[…] established with the assistance of the Hardy-Littlewood maximal inequality; see for instance this previous blog post. A remarkable observation of Stein, known as Stein’s maximal principle, allows one to reverse […]

10 August, 2020 at 11:53 am

Zijin LiuIn exercise 42, do you mean using sub-collection constructed in Vitali Covering lemma to cover the compact set K (with two times radius)? I have been stuck for a couple of days, could you provide some further hints?

Thank you.

10 August, 2020 at 7:42 pm

Terence TaoBasically yes, except that one has to use times the radius rather than twice the radius; also one has to use compactness more efficiently to make covered by the rather than by .

17 August, 2020 at 10:13 pm

Zijin LiuDear Professor Tao,

In exercise 58, I was trying to use Vitali-type covering lemma to modify the proof of Theorem 36. But I can only get an h_x for each x, such that [F(x+h_x)-F(x)/h_x]>λ, and (x,x+h_x) may not form an open cover, since it is not centered at x. Could you provide some hint to solve this issue?

Thank you

20 August, 2020 at 2:15 pm

Terence TaoOne can create an epsilon of room here and enlarge each by a tiny amount to generate an open cover without degrading the lower bound on the difference quotient by too much, and then take limits at the end of the argument to remove the loss.

19 February, 2021 at 9:03 am

N is a numberFor solving Exercise 16, I think one can use Fubini’s theorem: we have and so taking one more limit as and then interchanging the limits (thanks to Fubini’s theorem, since both of these exists and the integrand is absolutely integrable), the claim follows.

Is there a way to avoid using Fubini’s theorem (something which is easier than proof of Fubini’s theorem for instance).

[Dominated convergence works just fine here – T.]21 February, 2021 at 9:00 am

N is a numberIt can also be proved using density argument along the lines of proof of Proposition 19.